For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8
cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .
1 answer:
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Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So